Explanation:
Commercially available batteries use a variety of metals and electrolytes. Anodes can be made of zinc, aluminum, lithium, cadmium, iron, metallic lead, lanthanide, or graphite. Cathodes can be made of manganese dioxide, mercuric oxide, nickel oxyhydroxide, lead dioxide or lithium oxide. Potassium hydroxide is the electrolyte used in most battery types, but some batteries use ammonium or zinc chloride, thionyl chloride, sulfuric acid or lithiated metal oxides. The exact combination varies by battery type. For example, common single-use alkaline batteries use a zinc anode, a manganese dioxide cathode, and potassium hydroxide as the electrolyt
K = C + 273, so 27°C = 27+273 = 300 K
1 dg = 100 mg, so 20 dg = 20×100 = 2,000 mg
It is 29 and a half days long
Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F = 
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²
We have: Energy(E) = Planck's constant(h) × Frequency(∨)
Here, Planck's constant(h) = 6.626 × 10⁻³⁴ J/s
Frequency (∨) = 3.16 × 10¹² /s
Substitute the values into the expression:
E = (6.626 × 10⁻³⁴)(3.16 × 10¹²) J
E = 2.093 × 10⁻²¹ Joules
In short, Your Final answer would be 2.093 × 10⁻²¹ J
Hope this helps!
